Here are the solutions to the problems: 1. A body moving with a velocity of $15 \text{ m/s}$ is uniformly accelerated at $3 \text{ m/s}^2$. Calculate its velocity after travelling $20 \text{ m}$. Given: Initial velocity, $u = 15 \text{ m/s}$ Acceleration, $a = 3 \text{ m/s}^2$ Displacement, $s = 20 \text{ m}$ Final velocity, $v = ?$ Step 1: Use the kinematic equation $v^2 = u^2 + 2as$. $$v^2 = (15 \text{ m/s})^2 + 2(3 \text{ m/s}^2)(20 \text{ m})$$ $$v^2 = 225 \text{ m}^2/\text{s}^2 + 120 \text{ m}^2/\text{s}^2$$ $$v^2 = 345 \text{ m}^2/\text{s}^2$$ Step 2: Solve for $v$. $$v = \sqrt{345 \text{ m}^2/\text{s}^2}$$ $$v \approx 18.57 \text{ m/s}$$ The final velocity is $\boxed{\text{18.57 m/s}}$. --- 2. A particle starts from rest and covers $45 \text{ m}$ in $5 \text{ sec}$ under uniform acceleration. Determine the acceleration. Given: Initial velocity, $u = 0 \text{ m/s}$ (starts from rest) Displacement, $s = 45 \text{ m}$ Time, $t = 5 \text{ s}$ Acceleration, $a = ?$ Step 1: Use the kinematic equation $s = ut + \frac{1}{2}at^2$. $$45 \text{ m} = (0 \text{ m/s})(5 \text{ s}) + \frac{1}{2}a(5 \text{ s})^2$$ $$45 \text{ m} = 0 + \frac{1}{2}a(25 \text{ s}^2)$$ $$45 \text{ m} = 12.5a \text{ s}^2$$ Step 2: Solve for $a$. $$a = \frac{45 \text{ m}}{12.5 \text{ s}^2}$$ $$a = 3.6 \text{ m/s}^2$$ The acceleration is $\boxed{\text{3.6 m/s}^2}$. --- 3. A train travelling at $72 \text{ km/h}$ is brought to rest in $10 \text{ seconds}$. Determine: (i) the acceleration ($\text{m/s}^2$) (ii) the distance travelled before stopping ($\text{m}$) Given: Initial velocity, $u = 72 \text{ km/h}$ Final velocity, $v = 0 \text{ m/s}$ (brought to rest) Time, $t = 10 \text{ s}$ Step 1: Convert initial velocity from $\text{km/h}$ to $\text{m/s}$. $$u = 72 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}$$ $$u = 72 \times \frac{1000}{3600} \text{ m/s}$$ $$u = 20 \text{ m/s}$$ (i) Calculate the acceleration: Step 2: Use the kinematic equation $v = u + at$. $$0 \text{ m/s} = 20 \text{ m/s} + a(10 \text{ s})$$ $$-20 \text{ m/s} = 10a \text{ s}$$ $$a = \frac{-20 \text{ m/s}}{10 \text{ s}}$$ $$a = -2 \text{ m/s}^2$$ The acceleration is $\boxed{\text{-2 m/s}^2}$. (The negative sign indicates deceleration). (ii) Calculate the distance travelled before stopping: Step 3: Use the kinematic equation $s = ut + \frac{1}{2}at^2$. $$s = (20 \text{ m/s})(10 \text{ s}) + \frac{1}{2}(-2 \text{ m/s}^2)(10 \text{ s})^2$$ $$s = 200 \text{ m} - (1 \text{ m/s}^2)(100 \text{ s}^2)$$ $$s = 200 \text{ m} - 100 \text{ m}$$ $$s = 100 \text{ m}$$ Alternatively, using $s = \frac{u+v}{2}t$: $$s = \frac{20 \text{ m/s} + 0 \text{ m/s}}{2} \times 10 \text{ s}$$ $$s = \frac{20 \text{ m/s}}{2} \times 10 \text{ s}$$ $$s = 10 \text{ m/s} \times 10 \text{ s}$$ $$s = 100 \text{ m}$$ The distance travelled before stopping is $\boxed{\text{100 m}}$. --- 4. A body accelerates uniformly from $5 \text{ m/s}$ to $25 \text{ m/s}$ over a distance of $100 \text{ m}$. Calculate: (i) its acceleration (ii) time taken Given: Initial velocity, $u = 5 \text{ m/s}$ Final velocity, $v = 25 \text{ m/s}$ Displacement, $s = 100 \text{ m}$ (i) Calculate its acceleration: Step 1: Use the kinematic equation $v^2 = u^2 + 2as$. $$(25 \text{ m/s})^2 = (5 \text{ m/s})^2 + 2a(100 \text{ m})$$ $$625 \text{ m}^2/\text{s}^2 = 25 \text{ m}^2/\text{s}^2 + 200a \text{ m}$$ $$625 - 25 = 200a$$ $$600 = 200a$$ $$a = \frac{600}{200} \text{ m/s}^2$$ $$a = 3 \text{ m/s}^2$$ The acceleration is $\boxed{\text{3 m/s}^2}$. (ii) Calculate the time taken: Step 2: Use the kinematic equation $v = u + at$. $$25 \text{ m/s} = 5 \text{ m/s} + (3 \text{ m/s}^2)t$$ $$25 - 5 = 3t$$ $$20 = 3t$$ $$t = \frac{20}{3} \text{ s}$$ $$t \approx 6.67 \text{ s}$$ The time taken is $\boxed{\text{6.67 s}}$. --- 5. A particle moves such that its displacement is given by: $S = 4t + 2t^2 \text{ (m)}$. Determine: (i) its initial velocity (ii) its acceleration, and state its nature (iii) its velocity at $5 \text{ seconds}$ Given: Displacement $S = 4t + 2t^2$. Recall: Velocity $v = \frac{dS}{dt}$ Acceleration $a = \frac{dv}{dt}$ Step 1: Find the velocity function by differentiating $S$ with respect to $t$. $$v = \frac{d}{dt}(4t + 2t^2)$$ $$v = 4 + 4t \text{ m/s}$$ (i) Determine its initial velocity: Step 2: Initial velocity is the velocity at $t=0$. $$v(0) = 4 + 4(0)$$ $$v(0) = 4 \text{ m/s}$$ The initial velocity is $\boxed{\text{4 m/s}}$. (ii) Determine its acceleration, and state its nature: Step 3: Find the acceleration function by differentiating $v$ with respect to $t$. $$a = \frac{d}{dt}(4 + 4t)$$ $$a = 4 \text{ m/s}^2$$ The acceleration is $\boxed{\text{4 m/s}^2}$. Since the acceleration is a constant positive value, its nature is uniform acceleration. (iii) Determine its velocity at $5 \text{ seconds}$: Step 4: Substitute $t=5 \text{ s}$ into the velocity function. $$v(5) = 4 + 4(5)$$ $$v(5) = 4 + 20$$ $$v(5) = 24 \text{ m/s}$$ The velocity at $5 \text{ seconds}$ is $\boxed{\text{24 m/s}}$. --- 6. A car travelling at $15 \text{ m/s}$ accelerates uniformly at $1.2 \text{ m/s}^2$. How long will it take to reach $30 \text{ m/s}$? Given: Initial velocity, $u = 15 \text{ m/s}$ Acceleration, $a = 1.2 \text{ m/s}^2$ Final velocity, $v = 30 \text{ m/s}$ Time, $t = ?$ Step 1: Use the kinematic equation $v = u + at$. $$30 \text{ m/s} = 15 \text{ m/s} + (1.2 \text{ m/s}^2)t$$ $$30 - 15 = 1.2t$$ $$15 = 1.2t$$ $$t = \frac{15}{1.2} \text{ s}$$ $$t = 12.5 \text{ s}$$ It will take $\boxed{\text{12.5 s}}$. --- 7. A particle starts from rest and accelerates uniformly at $4 \text{ m/s}^2$ for $5 \text{ seconds}$ then immediately decelerates at $2 \text{ m/s}^2$ until it comes to rest. Find the total distance travelled by the particle. This problem has two phases: acceleration and deceleration. Phase 1: Acceleration Given: Initial velocity, $u_1 = 0 \text{ m/s}$ (starts from rest) Acceleration, $a_1 = 4 \text{ m/s}^2$ Time, $t_1 = 5 \text{ s}$ Step 1: Calculate the final velocity of Phase 1 ($v_1$). This will be the initial velocity for Phase 2. $$v_1 = u_1 + a_1t_1$$ $$v_1 = 0 \text{ m/s} + (4 \text{ m/s}^2)(5 \text{ s})$$ $$v_1 = 20 \text{ m/s}$$ Step 2: Calculate the distance travelled in Phase 1 ($s_1$). $$s_1 = u_1t_1 + \frac{1}{2}a_1t_1^2$$ $$s_1 = (0 \text{ m/s})(5 \text{ s}) + \frac{1}{2}(4 \text{ m/s}^2)(5 \text{ s})^2$$ $$s_1 = 0 + \frac{1}{2}(4 \text{ m/s}^2)(25 \text{ s}^2)$$ $$s_1 = 2 \text{ m/s}^2 \times 25 \text{ s}^2$$ $$s_1 = 50 \text{ m}$$ Phase 2: Deceleration Given: Initial velocity, $u_2 = v_1 = 20 \text{ m/s}$ Acceleration (deceleration), $a_2 = -2 \text{ m/s}^2$ Final velocity, $v_2 = 0 \text{ m/s}$ (comes to rest) Step 3: Calculate the distance travelled in Phase 2 ($s_2$). Use the kinematic equation $v_2^2 = u_2^2 + 2a_2s_2$. $$(0 \text{ m/s})^2 = (20 \text{ m/s})^2 + 2(-2 \text{ m/s}^2)s_2$$ $$0 = 400 \text{ m}^2/\text{s}^2 - 4s_2 \text{ m/s}^2$$ $$4s_2 \text{ m/s}^2 = 4